1 Answer
1

You could first make a substitution $u=\sqrt x$. The integral then becomes $\int_0^\infty 2u^{35} e^{-u}\,du$, which is easier to deal with. Here, towards making a comparison, show that for $u$ sufficiently large, $u^{35}<e^{ u/2}$.

For b):

Use the inequality $0\le \sin x\le x$ for $x\ge0$.

For c):

You could first make a substitution: $u=1-x^3$. This gives the integral
$$ {1\over3}\int_0^1 { du\over u^{1/3} (1-u)^{2/3} } =
{1\over3}\int_0^{1/2} { du\over u^{1/3} (1-u)^{2/3} }
+{1\over3}\int_{1/2}^1 { du\over u^{1/3} (1-u)^{2/3} }.$$
You should be able to deal with the latter two integrals (by essentially neglecting one of the terms in the denominator for each).